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A non-uniform bar of weight W is suspended at rest by two strings of negligible weight as shown in figure. The angles made by the strings with the vertical are 36.9^(@) and 53.1^(@) respectively. The bar is 2 m long. Calculate the distance d of the centre of gravity of the bar from its left end. |
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Answer» Solution :Suppose bar AB is suspended at rest by two strings OA and O.B of negligible weight. The angles made by the strings with the VERTICAL are `36.9^(@)and53.1^(@)` respectively. `angleOA A.=53.1^(@)andangleO.BB=36.9^(@)` `AB=2m` and suppose distance of centre of mass from end A is d.  `T_(1)andT_(2)` are the tensions produced in the left and right strings respectively and their mutual components are shown as in figure. At TRANSLATIONAL equilibrium, `-T_(1)cos53.1^(@)=-T_(2)cos36.9^(@)` `:.T_(1)cos53.1^(@)=T_(2)cos36.9^(@)and....(1)` `T_(1)sin53.1^(@)+T_(2)cos36.9^(@)=W....(2)` Sum of torque at point A `:.-T_(2)sin36.9^(@)xxAB+W.d=0` `:.-T_(2)sin36.9^(@)xx2+W.d=0` `:.T_(2)=(Wd)/(2sin36.9^(@))....(3)` From eqn. (2) and (3) `T_(1)sin53.1^(@)=W-T_(2)sin36.9^(@)` `=W-(Wdsin36.9^(@))/(2sin36.9^(@))` `=W-(Wd)/(2)` `:.T_(1)=(W(1-(d)/(2)))/(sin53.1^(@))....(4)` From eqn. (1) `T_(1)cos53.1^(@)=T_(2)cos36.9^(@)` PUTTING the VALUE of eqn. (2) and (3) `(W(1-(d)/(2))cos53.1^(@))/(sin53.1^(@))=(Wdcos36.9^(@))/(2sin36.9^(@))` `:.(1-(d)/(2))/(tan53.1^(@))=(d)/(2tan36.9^(@))` `:.(1-(d)/(2))/(1.3319)=(d)/(2xx0.7508)` `:.1-(d)/(2)=(d xx1.3319)/(1.5016)` `:.1=0.5d+0.887d` `:.1=1.387d` `:.d=(1)/(1.387)=0.72098` `:.d~~0.721m` `:.d~~72.1cm` |
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